29)

Derivation of equation (

In case of uniform scaling of velocity,

$$\begin{eqnarray} \frac{\partial z_{\tilde{\gamma}}}{\partial \rho_V} = -\frac{z_... ...-\gamma)} \frac{\partial S_r}{\partial \rho_{i}} \right),\nonumber\end{eqnarray}$$

$$\begin{eqnarray} \frac{\partial z_{\tilde{\gamma}}}{\partial \rho_V} = -\frac{z_... ...\gamma)} - \frac{S_r}{S_r\cos(\alpha_x-\gamma)} \right), \nonumber\end{eqnarray}$$

$$\begin{eqnarray} \frac{\partial z_{\tilde{\gamma}}}{\partial \rho_V} = \frac{z_\... ...{\cos(\alpha_x+\gamma)} + \frac{1}{\cos(\alpha_x-\gamma)} \right).\end{eqnarray}$$ (46)

The quantity $\frac{1}{\cos(\alpha_x+\gamma)} + \frac{1}{\cos(\alpha_x-\gamma)}$ can be written

$$\begin{eqnarray} \frac{1}{\cos(\alpha_x+\gamma)} + \frac{1}{\cos(\alpha_x-\gamma... ...)} &=&\frac{2\cos\alpha_x\cos\gamma}{\cos^2\alpha_x-\sin^2\gamma}.\end{eqnarray}$$ (47) (48) (49) (50)

Using equation (50), equation (46) simplifies to

$$\begin{eqnarray} \frac{\partial z_{\tilde{\gamma}}}{\partial \rho_V} &=& z_\xi \... ...2\alpha_x-\sin^2\gamma} \tan \gamma \tan \widehat{\gamma} \right).\end{eqnarray}$$ (51) (52) (53)